tag:blogger.com,1999:blog-72364578557859190452024-03-26T23:37:19.020-07:00Geometry Problems from IMOsolympiad geometry problems with aops links<br><br>geometry articles, books, magazines, shortlists for Juniors and Seniors
<br>problem collections with solutions from National, Regional and International Mathematical Olympiads
<br> latest added on the right menuparmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.comBlogger18125tag:blogger.com,1999:blog-7236457855785919045.post-28446170430130536122020-03-30T13:36:00.002-07:002020-03-30T15:55:09.961-07:00Kostas Dortsios' IMO 3D Geometry Solutions<div dir="ltr" style="text-align: left;" trbidi="on">
here I shall collect in one place links solutions to old problems from International Mathematical Olympiads, posted in this page all given by Kostas Dortsios [a Greek fond of 3D Geometry and New Technologies]<br />
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Soon there shall be a pdf collecting all the given solutions in one place</div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">1959 IMO Problem 6 (CZS) </span><span style="text-align: left;">(</span><a href="http://imogeometry.blogspot.gr/2017/07/1959-imo-problem-6-czs.html" style="text-align: left;" target="_blank">here </a><span style="text-align: left;">- Kostas Dortsios)</span><br />
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">Two planes, P and Q; intersect along the line p: The point A is given in the plane P; and the point C in the plane Q; neither of these points lies on the straight line p: Construct an isosceles trapezoid ABCD (with AB parallel to CD) in which a circle can be inscribed, and with vertices B and D lying in the planes P and Q respectively.</span></div>
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proposed by Czechoslovakia</div>
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<span style="font-family: "times new roman";">1959 IMO Problem 6 (CZS) [variation] </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;"> </span><span style="text-align: left;">(</span><a href="http://imogeometry.blogspot.gr/2017/07/1959-imo-problem-6-czs.html" style="text-align: left;" target="_blank">here </a><span style="text-align: left;">- Kostas Dortsios)</span></div>
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<span style="font-family: "times new roman";">[<b>variation </b>by Kostas Dortsios, ABDC instead of ABCD]</span></div>
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<span style="font-family: "times new roman";">Two planes, P and Q, intersect along the line p. The point A is given in the plane P, and the point C in the plane Q, neither of these points lies on the straight line p. Construct an isosceles trapezoid ABDC (with AB parallel to CD) in which a circle can be inscribed, and with vertices B and D lying in the planes P and Q respectively</span></div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">1960 IMO Problem 5 (CZS) </span><span style="text-align: left;">(</span><a href="https://imogeometry.blogspot.gr/2017/07/1960-imo-problem-5-czs.html" style="text-align: left;" target="_blank">here </a><span style="text-align: left;">- Kostas Dortsios) [only in Greek <span style="color: red;">at the moment</span>]</span><br />
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">Consider the cube ABCDA'B'C'D' </span><span lang="EN-US" style="font-family: "cmr12"; font-size: 12pt;">(with face </span><span lang="EN-US" style="font-family: "cmmi12"; font-size: 12pt;">ABCD<i> </i></span><span lang="EN-US" style="font-family: "cmr12"; font-size: 12pt;">directly above face </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">A'B'C'D'</span><span lang="EN-US" style="font-family: "cmr12"; font-size: 12pt;">).</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"></span></div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">(a) Find the locus of the midpoints of segments XY , where X is any point of AC and<span style="mso-spacerun: yes;"> </span>Y is any point of B'D'.</span></div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">(b) Find the locus of points Z which lie on the segments XY of part (a) such that <span style="mso-spacerun: yes;"> </span></span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; position: relative; top: 2pt;">ΖΥ = 2 ΧΖ</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;"><span style="mso-spacerun: yes;"> </span>.</span><br />
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proposed by Czechoslovakia<br />
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1961 IMO Problem 6 (ROM) (<span style="color: red;">soon</span>)</div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">Consider a plane ε <span style="mso-spacerun: yes;"> </span>and three non-collinear points A, B, C on the same side of ε, suppose the plane determined by these three points is not parallel to ε. In plane a take three arbitrary points A</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">, B</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">, C</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">. Let L, M, N be the midpoints of segments AA</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">, BB</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">, CC</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">, let G be the centroid of triangle LMN. (We will not consider positions of the points A</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">, B</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">, C</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">such that the points L, M, N do not form a triangle.) What is the locus of point G as A</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">, B</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">, C</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">. range independently over the plane </span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">ε</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;"> <span lang="EN-US">?</span></span></div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"><br /></span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">by Gheorghe D. Simionescu</span></div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif";">1962 IMO Problem 3 (CZS) </span><span style="text-align: left;">(<span style="color: red;">soon</span>)</span><span style="font-family: "times new roman" , serif;"> </span><br />
<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">Consider the cube ABCDA</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">B</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">C</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">D</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> (ABCD and A</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">B</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">C</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">D</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> are the upper and lower bases, respectively, and edges AA</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">, BB</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">, CC</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">, DD</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> are parallel). The point X moves at constant speed along the perimeter of the square ABCD in the direction ABCDA, and the point Y moves at the same rate along the perimeter of the square B</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">C</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">CB in the direction B</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">C</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">CB</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">Β</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">. Points X and Y begin their motion at the same instant from the starting positions A and B</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">΄</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">, respectively. Determine and draw the locus of the midpoints of the segments XY.</span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-77913957374188817402020-03-30T13:21:00.001-07:002020-03-30T13:37:36.928-07:00old IMO Plane Geometry Solutions<div dir="ltr" style="text-align: left;" trbidi="on">
here I shall collect in one place links solutions to old plane geometry problems from International Mathematical Olympiads, posted in this page<br />
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1959 IMO Problem 4 (HUN) (<a href="http://imogeometry.blogspot.gr/2017/07/1959-imo-problem-4-hun.html" target="_blank">here </a>- mine)<br />
Construct a right triangle with given hypotenuse c such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.<br />
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proposed by Hungary</div>
1959 IMO Problem 5 (ROM) (<a href="http://imogeometry.blogspot.gr/2017/07/1959-imo-problem-5-rom.html" target="_blank">here </a>- mine & Theodosis Giannopoulos)<br />
<span style="font-family: "cmr12"; font-size: 16px;">An arbitrary point M is selected in the interior of the segment AB. The squares AMCD and MBEF are constructed on the same side of AB; with the segments AM and MB as their respective bases. The circles circumscribed about these squares, with centers P and Q; intersect at M and also at another point N: Let N' denote the point of intersection of the straight lines AF and BC:</span><br />
<span style="font-family: "cmr12"; font-size: 16px;">(a) Prove that the points N and N' coincide.</span><br />
<span style="font-family: "cmr12"; font-size: 16px;">(b) Prove that the straight lines MN pass through a fixed point S independent of the choice of M.</span><br />
<span style="font-family: "cmr12"; font-size: 16px;">(c) Find the locus of the midpoints of the segments PQ as M varies between A and B.</span><br />
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proposed by Cezar Cosnita, Romania</div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-220442859524415942019-08-21T09:45:00.002-07:002019-08-21T09:49:03.249-07:002010 JBMO Shortlist G1 <div dir="ltr" style="text-align: left;" trbidi="on">
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<span lang="EN-US" style="font-family: "times new roman", serif; font-size: 12pt; text-align: justify;">Consider a triangle ${ABC}$</span><span lang="EN-US" style="font-family: "times new roman", serif; font-size: 12pt; position: relative; text-align: justify; top: 3pt;"> </span><span lang="EN-US" style="font-family: "times new roman", serif; font-size: 12pt; text-align: justify;">with${\angle ACB=90^{\circ}.}$ Let ${F}$ be the foot of the altitude from <span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;">${C}$</span>. Circle ${\omega}$ touches the line segment ${FB}$at point ${P,}$<span style="mso-spacerun: yes;"> </span>the altitude ${CF}$at point ${Q}$ and the circumcircle of ${ABC}$at point ${R.}$ Prove that points ${A,Q,R}$ are collinear and ${AP=AC}$.</span></div>
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<span style="color: red;">posted in aops <span style="font-size: large;"><a href="http://artofproblemsolving.com/community/c6h622420p3722136" target="_blank">here</a></span></span></div>
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<span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;"><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">my s</span><span style="font-family: "times new roman" , serif; font-size: 12pt;">olutio</span><span style="font-family: "times new roman" , serif; font-size: 12pt;">n</span></span><br />
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-76432750702977552622019-08-21T05:49:00.001-07:002020-03-30T13:12:24.381-07:00my JBMO Shortlist solutions<div dir="ltr" style="text-align: left;" trbidi="on">
here I shall collect in one place links to my solutions from Junior Balkan Math Olympiad Shortlist<br />
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<span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">2010 JBMO Shortlist G1 </span><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">(</span><a href="https://imogeometry.blogspot.com/2019/08/2010-jbmo-shortlist-g1.html" target="_blank">here</a><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">)</span></div>
<span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">Consider a triangle ${ABC}$</span><span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; position: relative; text-align: justify; top: 3pt;"> </span><span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">with${\angle ACB=90^{\circ}.}$ Let ${F}$ be the foot of the altitude from <span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;">${C}$</span>. Circle ${\omega}$ touches the line segment ${FB}$at point ${P,}$<span style="mso-spacerun: yes;"> </span>the altitude ${CF}$at point ${Q}$ and the circumcircle of ${ABC}$at point ${R.}$ Prove that points ${A,Q,R}$ are collinear and ${AP=AC}$.</span></div>
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<span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">2011 JBMO Shortlist G1 </span><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">(</span><a href="https://imogeometry.blogspot.com/2019/08/2011-jbmo-shortlist-g1.html" target="_blank">here</a><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">)</span><br />
<span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">Let $ABC$ be an isosceles triangle with $AB=AC$. On the extension of the side </span><span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;"><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">${CA}$</span></span> we consider the point ${D}$<i> </i>such that ${AD<AC}$. The perpendicular bisector of the segment ${BD}$ meets the internal and the external bisectors of the angle $\angle BAC$<i> </i>at the points ${E}$ and ${Z}$, respectively. Prove that the points ${A, E, D, Z}$<i> </i>are concyclic. </span></div>
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<span style="font-family: "times new roman" , serif;">2011 JBMO Shortlist G2 </span><span style="font-family: "times new roman" , serif; font-size: 16px;">(</span><a href="https://imogeometry.blogspot.com/2019/08/2011-jbmo-shortlist-g2.html" target="_blank">here</a><span style="font-family: "times new roman" , serif; font-size: 16px;">)</span></div>
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<span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">Let $AD,BF$ and ${CE}$ be the altitudes of $\vartriangle ABC$. A line passing through ${D}$ and parallel to ${AB}$intersects the line ${EF}$at the point ${G}$. If ${H}$ is the orthocenter of $\vartriangle ABC$, find the angle ${\angle{CGH}}$.</span><br />
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<span style="font-family: "times new roman" , serif;">2011 JBMO Shortlist G3 </span><span style="font-family: "times new roman" , serif; font-size: 16px;">(</span><a href="https://imogeometry.blogspot.com/2019/08/2011-jbmo-shortlist-g3.html" style="font-family: "times new roman", serif; font-size: 16px;" target="_blank">here</a><span style="font-family: "times new roman" , serif; font-size: 16px;">)</span></div>
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<span style="color: black; font-family: "times new roman" , serif; font-size: 12pt; font-style: normal; font-weight: 400; letter-spacing: normal; text-align: justify; text-transform: none; white-space: normal; word-spacing: 0px;">Let $ABC$ be a triangle in which (${BL}$is the angle bisector of ${\angle{ABC}}$ $\left( L\in AC \right)$, ${AH}$ is an altitude of$\vartriangle ABC$ $\left( H\in BC \right)$ and ${M}$is the midpoint of the side ${AB}$. It is known</span><span style="color: black; font-family: "times new roman" , serif; font-size: 12pt; font-style: normal; font-weight: 400; letter-spacing: normal; text-align: justify; text-transform: none; white-space: normal; word-spacing: 0px;"> </span><span style="color: black; font-family: "times new roman" , serif; font-size: 12pt; font-style: normal; font-weight: 400; letter-spacing: normal; text-align: justify; text-transform: none; white-space: normal; word-spacing: 0px;">that the midpoints of the segments ${BL}$ and ${MH}$ coincides. Determine the internal angles of triangle $\vartriangle ABC$. </span><br />
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<span style="font-family: "times new roman" , serif;">2012 JBMO Shortlist G2 </span><span style="font-family: "times new roman" , serif; font-size: 16px;">(</span><a href="https://imogeometry.blogspot.com/2019/08/2012-jbmo-shortlist-g2.html" target="_blank">here</a><span style="font-family: "times new roman" , serif; font-size: 16px;">)</span><br />
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<span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; line-height: 18.4px;">Let $ABC$ be an isosceles triangle with $AB=AC$. Let also ${c\left(K, KC\right)}$ be a circle tangent to the line ${AC}$</span><span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; line-height: 18.4px; position: relative; top: 3pt;"> </span><span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; line-height: 18.4px;">at point${C}$ which it intersects the segment ${BC}$ again at an interior point ${H}$. Prove that ${HK\perp AB}$.</span></div>
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<span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">2013 JBMO Shortlist G1 </span><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">(</span><a href="https://imogeometry.blogspot.com/2019/08/2013-jbmo-shortlist-g1.html" target="_blank">here</a><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">)</span><br />
<span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">Let ${AB}$ be a diameter of a circle</span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;"> </span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">${\omega}$ and center ${O}$ ,</span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;"> </span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;"> </span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">${OC}$ a radius of ${\omega}$ perpendicular to $AB$,${M}$ be a point of the segment $\left( OC \right)$ . Let ${N}$ be the second intersection point of line ${AM}$ with ${\omega}$ and ${P}$ the intersection point of the tangents of ${\omega}$ at points ${N}$ and ${B.}$ Prove that points ${M,O,P,N}$ are cocyclic.</span><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;"></span><br />
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<span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">2014 JBMO Shortlist G1 </span><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">(</span><a href="https://imogeometry.blogspot.com/2017/09/2014-jbmo-shortlist-g1.html" target="_blank">here</a><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">)</span></div>
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<span style="text-align: justify;">Let $ABC$ be a triangle with $m\left( \angle B \right)=m\left( \angle C \right)={{40}^{{}^\circ }}$ Line bisector of ${\angle{B}}$ intersects ${AC}$ at point ${D}$. Prove that $BD+DA=BC$.</span></div>
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<span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">2015 JBMO Shortlist G1 </span><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">(</span><a href="https://imogeometry.blogspot.com/2019/08/2015-jbmo-shortlist-g1.html" target="_blank">here</a><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">)</span></div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">Around the triangle $ABC$ the circle is circumscribed, and at the vertex ${C}$</span><span lang="EN-US" style="position: relative; text-align: justify; top: 3pt;"> </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">tangent ${t}$ to this circle is drawn. The line ${p}$, which is parallel to this tangent intersects the lines ${BC}$ and ${AC}$ at the points ${D}$ and ${E}$, respectively. Prove that the points $A,B,D,E$ belong to the same circle.</span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-6604750065364917312019-08-09T14:40:00.002-07:002019-08-21T05:54:58.449-07:002011 JBMO Shortlist G3<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="font-family: "times new roman", serif; font-size: 12pt; text-align: justify;">Let $ABC$ be a triangle in which (${BL}$is the angle bisector of ${\angle{ABC}}$ $\left( L\in AC \right)$, ${AH}$ is an altitude of$\vartriangle ABC$ $\left( H\in BC \right)$ and ${M}$is the midpoint of the side ${AB}$. It is known</span><span style="font-family: "times new roman", serif; font-size: 12pt; text-align: justify;"> </span><span style="font-family: "times new roman", serif; font-size: 12pt; text-align: justify;">that the midpoints of the segments ${BL}$ and ${MH}$ coincides. Determine the internal angles of triangle $\vartriangle ABC$.</span></div>
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<span style="color: red;">posted in aops <span style="font-size: large;"><a href="http://artofproblemsolving.com/community/c6h1525209p9137512" target="_blank">here</a></span></span></div>
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<span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;"><span style="font-family: "times new roman" , serif; font-size: 16px; text-align: justify;">my s</span><span style="font-family: "times new roman" , serif; font-size: 12pt;">olutio</span><span style="font-family: "times new roman" , serif; font-size: 12pt;">n </span></span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-69049544890167932502019-08-07T09:38:00.002-07:002019-08-07T09:52:09.351-07:002011 JBMO Shortlist G1<div dir="ltr" style="text-align: left;" trbidi="on">
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<span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">Let $ABC$ be an isosceles triangle with $AB=AC$. On the extension of the side </span><span lang="EN-US" style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;"><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">${CA}$</span></span> we consider the point ${D}$<i> </i>such that ${AD<AC}$. The perpendicular bisector of the segment ${BD}$ meets the internal and the external bisectors of the angle $\angle BAC$<i> </i>at the points ${E}$ and ${Z}$, respectively. Prove that the points ${A, E, D, Z}$<i> </i>are concyclic.</span></div>
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<span style="color: red;">posted in aops <span style="font-size: large;"><a href="http://artofproblemsolving.com/community/c6h1525207p9137502" target="_blank">here</a></span></span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-57860984207068930832019-08-06T18:57:00.002-07:002019-08-06T19:07:37.361-07:002011 JBMO Shortlist G2<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">Let $AD,BF$ and ${CE}$ be the altitudes of $\vartriangle ABC$. A line passing through ${D}$ and parallel to ${AB}$intersects the line ${EF}$at the point ${G}$. If ${H}$ is the orthocenter of $\vartriangle ABC$, find the angle ${\angle{CGH}}$.</span></div>
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<span style="color: red;">posted in aops <span style="font-size: large;"><a href="http://artofproblemsolving.com/community/c6h1525208p9137505" target="_blank">here</a></span></span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-65611856540388341632019-08-04T17:28:00.001-07:002019-08-04T17:41:59.226-07:002012 JBMO Shortlist G2<div dir="ltr" style="text-align: left;" trbidi="on">
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<span lang="EN-US" style="font-family: "times new roman", serif; font-size: 12pt; line-height: 18.4px;">Let $ABC$ be an isosceles triangle with $AB=AC$. Let also ${c\left(K, KC\right)}$ be a circle tangent to the line ${AC}$</span><span lang="EN-US" style="font-family: "times new roman", serif; font-size: 12pt; line-height: 18.4px; position: relative; top: 3pt;"> </span><span lang="EN-US" style="font-family: "times new roman", serif; font-size: 12pt; line-height: 18.4px;">at point${C}$ which it intersects the segment ${BC}$ again at an interior point ${H}$. Prove that ${HK\perp AB}$.</span></div>
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<span style="color: red;">posted in aops <span style="font-size: large;"><a href="https://artofproblemsolving.com/community/c6h623857p3734395" target="_blank">here</a></span></span></div>
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<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;">my solution </span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"><b style="font-family: "times new roman", serif;">without words</b></span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"><b style="font-family: "times new roman", serif;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgy0oOmcMKVdSXvSg6A04be9WO_DNCfhaBUxU2vpjxKI_lr09XENYQmdeOdIX0U8sy3u5pORCB0loKPhYYaprwr96_EdmK6rycG8xxYLh4oPVKfdyf2yTpqzgD7k0EYLUtbxh6sbQ4R_ndQ/s1600/2012+JBMO+Shortlist+G2.png" imageanchor="1"><img border="0" height="471" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgy0oOmcMKVdSXvSg6A04be9WO_DNCfhaBUxU2vpjxKI_lr09XENYQmdeOdIX0U8sy3u5pORCB0loKPhYYaprwr96_EdmK6rycG8xxYLh4oPVKfdyf2yTpqzgD7k0EYLUtbxh6sbQ4R_ndQ/s640/2012+JBMO+Shortlist+G2.png" width="640" /></a></b></span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-71441072052493534532019-08-01T11:32:00.001-07:002019-08-04T17:14:43.190-07:002015 JBMO Shortlist G1<div dir="ltr" style="text-align: left;" trbidi="on">
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">Around the triangle $ABC$ the circle is circumscribed, and at the vertex ${C}$</span><span lang="EN-US" style="position: relative; text-align: justify; top: 3pt;"> </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt; text-align: justify;">tangent ${t}$ to this circle is drawn. The line ${p}$, which is parallel to this tangent intersects the lines ${BC}$ and ${AC}$ at the points ${D}$ and ${E}$, respectively. Prove that the points $A,B,D,E$ belong to the same circle.</span><br />
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<span style="color: red;"></span>
<span style="color: red;">posted in aops <span style="font-size: large;"><a href="http://artofproblemsolving.com/community/c6h1525168p9137006" target="_blank">here</a></span></span></div>
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<span style="font-family: "times new roman" , serif; font-size: 12pt;">my solution </span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"><b style="font-family: "times new roman", serif;">without words</b></span><br />
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<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"><b style="font-family: "times new roman", serif;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWWCTt8h-ohMopo5z8CiKnt918v-yTfb3kvdYZwIh_89oifCFe4FVRdHOaQWO0A0m892nIbYUeZcYW3R8kV20uCpGBBZfjV2lqKwWljHWgIxXPiJcjocPGEmQDLTfD_yqa_5H9zRau8c3i/s1600/2015+JBMO+Shortlist+G1.png" imageanchor="1"><img border="0" height="450" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiWWCTt8h-ohMopo5z8CiKnt918v-yTfb3kvdYZwIh_89oifCFe4FVRdHOaQWO0A0m892nIbYUeZcYW3R8kV20uCpGBBZfjV2lqKwWljHWgIxXPiJcjocPGEmQDLTfD_yqa_5H9zRau8c3i/s640/2015+JBMO+Shortlist+G1.png" width="640" /></a></b></span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-84932214040381209742019-08-01T01:49:00.002-07:002019-08-04T17:17:13.233-07:002013 JBMO Shortlist G1<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">Let ${AB}$ be a diameter of a circle</span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;"> </span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">${\omega}$ and center ${O}$ ,</span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;"> </span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;"> </span><span style="font-family: "times new roman" , serif; font-size: 12pt; text-align: justify;">${OC}$ a radius of ${\omega}$ perpendicular to $AB$,${M}$ be a point of the segment $\left( OC \right)$ . Let ${N}$ be the second intersection point of line ${AM}$ with ${\omega}$ and ${P}$ the intersection point of the tangents of ${\omega}$ at points ${N}$ and ${B.}$ Prove that points ${M,O,P,N}$ are cocyclic</span></div>
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<span style="color: red;">posted in aops <span style="font-size: large;"><a href="http://artofproblemsolving.com/community/c6h1525198p9137398" target="_blank">here</a></span></span></div>
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<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;">my solution </span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"><b style="font-family: "times new roman", serif;">without words</b></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiI6OxtTXXOYbCxCiUT7s1ac5fmiAn5doDBlxfpumgKTrV_FNR6zABD3S8JXAEuwb49pRGcq_EIoATpOC17SSutAvbREtx5lrK9Pz803X_Ucdn650NdB_X8j8TBScINBooAlUJgepIQHwEg/s1600/2013+JBMO+Shortlist+G1.png" imageanchor="1"><img border="0" height="424" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiI6OxtTXXOYbCxCiUT7s1ac5fmiAn5doDBlxfpumgKTrV_FNR6zABD3S8JXAEuwb49pRGcq_EIoATpOC17SSutAvbREtx5lrK9Pz803X_Ucdn650NdB_X8j8TBScINBooAlUJgepIQHwEg/s640/2013+JBMO+Shortlist+G1.png" width="640" /></a></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-88347997724090514652018-09-01T18:37:00.000-07:002018-09-01T18:37:01.781-07:001985 BMO Shortlist 2 (GRE)<div dir="ltr" style="text-align: left;" trbidi="on">
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif;"><span style="font-size: 13px;">proposed by Dimitris Kontogiannis, Greece</span></span></div>
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif;"><span style="font-size: 13px;">[solved, alternate solutions are welcome]</span></span></div>
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif;"><span style="font-size: 13px;">Let ABC be a triangle with < A =120ο. Let AD, CE be the angle bisectors of angles < Α, < C respectively and I be the intersection point of AD, CE. If Z is the intersection point of BI and DE, calculate angle < DΑZ .</span></span></div>
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif;"><span style="font-size: 13px;"> posted in facebook </span><span style="font-size: large;"><a href="https://web.facebook.com/groups/parmenides52/permalink/1385271498253219/" target="_blank">here </a></span></span></div>
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif;"><span style="font-size: 13px;">Έστω τρίγωνο ABC με <A =120ο . Ας είναι AD, CE οι διχοτόμοι των γωνιών < Α, < C αντίστοιχα και I το σημείο τομής των AD, CE. Αν Z είναι το σημείο τομής των BI και DE , να υπολογίσετε την γωνία < DΑZ .</span></span></div>
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif;"><span style="font-size: 13px;"> solved by Leo Giugiuc</span></span></div>
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<span style="color: #222222; font-family: Arial, Helvetica, sans-serif;"><span style="font-size: 13px;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXoGFJDfIfqyHpPYx6TYbCZZlrhXT77tAsXXdVddRMsZpWhjrY_9Hyhe4AllGegE0_xPcJKtZIqN1Cb4mmEd96T4QXpjmfetcE9aycddtOQZYuNzbHCYpZMuKEmzwwgLbb4Gb8KuJf-S0y/s1600/BMO85+SL2+Leo.jpg" imageanchor="1"><img border="0" height="210" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXoGFJDfIfqyHpPYx6TYbCZZlrhXT77tAsXXdVddRMsZpWhjrY_9Hyhe4AllGegE0_xPcJKtZIqN1Cb4mmEd96T4QXpjmfetcE9aycddtOQZYuNzbHCYpZMuKEmzwwgLbb4Gb8KuJf-S0y/s640/BMO85+SL2+Leo.jpg" width="640" /></a></span></span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-10231688952266170342017-09-02T04:40:00.004-07:002019-08-17T06:54:22.786-07:002014 JBMO Shortlist G1 <div dir="ltr" style="text-align: left;" trbidi="on">
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Let $ABC$ be a triangle with $m\left( \angle B \right)=m\left( \angle C \right)={{40}^{{}^\circ }}$ Line bisector of ${\angle{B}}$ intersects ${AC}$ at point ${D}$. Prove that $BD+DA=BC$.</div>
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<span style="color: red;">posted in aops <span style="font-size: large;"><a href="http://artofproblemsolving.com/community/c6h1525185p9137218" target="_blank">here</a></span></span><br />
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<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;">Έστω</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;">τρίγωνο</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"> <span lang="EN-US">$\vartriangle ABC$ </span></span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;">με</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"> <span lang="EN-US">${\angle{B}=\angle{C}=40^{\circ}}$. </span></span><span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;">Η διχοτόμος της γωνίας ${\angle{B}}$ τέμνει την ${AC}$ στο σημείο ${D}$. Να αποδείξετε ότι $BD+DA=BC$.</span><br />
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<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;">my solution </span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"><b>without words</b></span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12pt; line-height: 18.4px;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9dwXOOVPI-MFPcsYOlGuwb3H6posQ9JqFmjwa9FOZJAeQlFj8SfshpBRX9lN6AzXsrCPh23VN4g1_Dl1qCIqaXQXmEmCEohjP8Fd5Tfj-0OKt58vX8_RWJDNqOHG5DkmtXkBrcVq5UkSO/s1600/jbmo+2014+g1.png" imageanchor="1"><img border="0" height="307" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi9dwXOOVPI-MFPcsYOlGuwb3H6posQ9JqFmjwa9FOZJAeQlFj8SfshpBRX9lN6AzXsrCPh23VN4g1_Dl1qCIqaXQXmEmCEohjP8Fd5Tfj-0OKt58vX8_RWJDNqOHG5DkmtXkBrcVq5UkSO/s640/jbmo+2014+g1.png" width="640" /></a></span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-34253264702417909492017-09-02T02:06:00.001-07:002017-10-16T09:32:18.490-07:001985 BMO Shortlist 1 (GRE)<div dir="ltr" style="text-align: left;" trbidi="on">
proposed by Theodoros Bolis, Greece<br />
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">Let e<sub>1</sub>, e<sub>2</sub> be two lines </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">perpendicular </span>to the same plane. Find the locus of the points of the space, that we can draw 3 lines, perpendicular in pairs, who intersect e<sub>1</sub> or e<sub>2 </sub>.</span></div>
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<span style="font-family: "times new roman" , "serif"; font-size: 12pt;">Ας είναι </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> e<sub>1</sub>, e<sub>2</sub> </span><sub><span style="font-family: "times new roman" , "serif"; font-size: 12pt;"></span></sub><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">δυο ευθείες κάθετες στο ίδιο επίπεδο.<span style="mso-spacerun: yes;"> </span>Να βρείτε τον γεωμετρικό τόπο των σημείων του χώρου από τα οποία μπορούμε να φέρουμε τρεις ευθείες ανα δυο κάθετες που να τέμνουν την </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">e<sub>1</sub></span><sub><span style="font-family: "times new roman" , "serif"; font-size: 12pt;"><span style="mso-spacerun: yes;"> </span></span></sub><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">είτε την </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">e<sub>2 </sub></span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">.</span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-35470023143210534852017-09-02T02:04:00.001-07:002018-06-28T00:12:31.586-07:001984 BMO Problem 2 (ROM)<div dir="ltr" style="text-align: left;" trbidi="on">
proposed by Romania<br />
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[solved , alternate solutions are welcome]</div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">Let <span style="mso-no-proof: yes;">ABCD </span>be a cyclic quadrilateral and let <span style="mso-no-proof: yes;">H<sub>A</sub>, H<sub>B</sub>, H<sub>C</sub>, H<sub>D</sub> <span style="mso-spacerun: yes;"> </span></span>be the orthocenters of the triangles BCD, CDA, DAB and ABC respectively. Show that the quadrilaterals <span style="mso-no-proof: yes;">ABCD </span>and <span style="mso-no-proof: yes;">H<sub>A</sub>H<sub>B</sub>H<sub>C</sub>H<sub>D</sub> <span style="mso-spacerun: yes;"> </span></span>are congruent.</span></div>
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<span style="color: red;"><span style="color: red;">posted </span>in facebook </span> <a href="https://web.facebook.com/groups/parmenides52/permalink/1385052191608483/"><span style="font-size: large;">here</span></a><br />
<span style="font-size: large;"><span style="font-size: small;">solved in aops</span> <a href="https://artofproblemsolving.com/community/c6h145339p822803" target="_blank">here</a></span></div>
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<span style="font-family: "times new roman" , "serif"; font-size: 12pt;">Θεωρούμε εγγράψιμο τετράπλευρο </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">ABCD</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">και τα ορθόκεντρα<span style="mso-no-proof: yes;"> </span></span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">H<sub>A</sub></span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">, </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">H<sub>B</sub></span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">, </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">H<sub>C</sub></span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">, </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">H<sub>D</sub></span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> <span style="mso-spacerun: yes;"> </span></span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">των τριγώνων</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;"> </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">BCD</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">, </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">CDA</span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">, </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">DAB</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">και </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">ABC</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">αντίστοιχως. Να αποδείξετε οτι τα τετράπλευρα </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">ABCD</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">and</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;">H<sub>A</sub>H<sub>B</sub>H<sub>C</sub>H<sub>D</sub></span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12pt;"> <span style="mso-spacerun: yes;"> </span></span><span style="font-family: "times new roman" , "serif"; font-size: 12pt;">είναι ίσα.</span><br />
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solved by Leo Giugiuc (using complex numbers)<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMsT3872LnIFMVHnyhyphenhyphenjDy-utbDfD5KPeyWpC64lZeSbKn3Zd_PSL1EvZsu3vs6fn1NUMSAe4Vek9WbsesgsEA_JCc95r_gmd6QrLF4gjYDRFRYZIR7FDr3ykSyr_rGYG4KncL5zIbwVsE/s1600/bmo+leo.jpg" imageanchor="1"><img border="0" height="228" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjMsT3872LnIFMVHnyhyphenhyphenjDy-utbDfD5KPeyWpC64lZeSbKn3Zd_PSL1EvZsu3vs6fn1NUMSAe4Vek9WbsesgsEA_JCc95r_gmd6QrLF4gjYDRFRYZIR7FDr3ykSyr_rGYG4KncL5zIbwVsE/s640/bmo+leo.jpg" width="640" /></a></div>
parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-4892023657315632412017-07-25T17:06:00.001-07:002017-10-16T09:10:55.322-07:001960 IMO Problem 5 (CZS)<div dir="ltr" style="text-align: left;" trbidi="on">
proposed by Czechoslovakia<br />
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[solved, alternate solutions are welcome]</div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Consider the cube ABCDA'B'C'D' </span><span lang="EN-US" style="font-family: "cmr12"; font-size: 12.0pt;">(with
face </span><span lang="EN-US" style="font-family: "cmmi12"; font-size: 12.0pt;">ABCD<i>
</i></span><span lang="EN-US" style="font-family: "cmr12"; font-size: 12.0pt;">directly above face </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">A'B'C'D'</span><span lang="EN-US" style="font-family: "cmr12"; font-size: 12.0pt;">).</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"></span></div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">(a) Find the locus of the
midpoints of segments XY , where X is any point of AC and<span style="mso-spacerun: yes;"> </span>Y is any point of B'D'.</span></div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">(b) Find the locus of points Z
which lie on the segments XY of part (a) such that<span style="mso-spacerun: yes;"> </span></span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt; position: relative; top: 2.0pt;">ΖΥ =
2 ΧΖ</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"><span style="mso-spacerun: yes;"> </span>.</span></div>
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<span style="font-family: "times new roman", serif; font-size: small;">solved in aops </span><a href="https://artofproblemsolving.com/community/c6h54831p341557" style="font-family: "times new roman", serif; font-size: x-large;" target="_blank">here</a></div>
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<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Δίνεται ο κύβος<span style="mso-spacerun: yes;"> </span></span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">ABCDA</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">'</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">B</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">'</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">C</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">'</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">D</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">' </span><span style="font-family: "cmr12"; font-size: 12.0pt;">(</span><span style="font-size: 12.0pt; mso-bidi-font-family: CMR12;">με την πλευρά</span><span style="font-family: "cmr12"; font-size: 12.0pt;"> </span><span lang="EN-US" style="font-family: "cmmi12"; font-size: 12.0pt;">ABCD</span><i><span lang="EN-US" style="font-family: "cmmi12"; font-size: 12.0pt;"> </span></i><span style="font-size: 12.0pt; mso-bidi-font-family: CMR12;">ακριβώς</span><span style="font-family: "cmr12"; font-size: 12.0pt;"> </span><span style="font-size: 12.0pt; mso-bidi-font-family: CMR12;">πάνω από την </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">A</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">'</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">B</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">'</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">C</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">'</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">D</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">'</span><span style="font-family: "cmr12"; font-size: 12.0pt;">).</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"></span></div>
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<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">(</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">a</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">) Βρείτε τον
γεωμετρικό τόπο των μέσων των τμημάτων </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">XY</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> , όπου </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">X</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">είναι<span style="mso-spacerun: yes;"> </span>οποιοδήποτε σημείο του <span style="mso-spacerun: yes;"> </span></span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">AC</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">και </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Y</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">οποιοδήποτε σημείο του </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">B</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">'</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">D</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">'.</span></div>
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<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">(</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">b</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">) Βρείτε τον
γεωμετρικό τόπο των σημείων </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Z</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">που ανήκουν στα τμήματα </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">XY</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">του<span style="mso-spacerun: yes;"> </span>ερωτήματος (</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">a</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">) , τέτοια ώστε <span style="mso-spacerun: yes;"> </span><span style="mso-text-raise: -2.0pt; position: relative; top: 2.0pt;">ΖΥ = 2 ΧΖ</span><span style="mso-spacerun: yes;"> </span>.</span><br />
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<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"></span><br />
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<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">solution by Kostas Dortsios in Greek </span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">1/2</span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> <a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJvSpFXo_yO_T6qKDL5uyZb9ybjcQGQ5XEEgMZadLPho7gos3UEm1cxm6BXw3LPo2Lw1G8cDO9C5NYEQj4ps4glm_N6VPiK3_b2EWNO5HliipOhhk06P91FWYpb6CVpLADDkYehm5Z76wc/s1600/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+1960_Page_1.png" imageanchor="1"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiJvSpFXo_yO_T6qKDL5uyZb9ybjcQGQ5XEEgMZadLPho7gos3UEm1cxm6BXw3LPo2Lw1G8cDO9C5NYEQj4ps4glm_N6VPiK3_b2EWNO5HliipOhhk06P91FWYpb6CVpLADDkYehm5Z76wc/s640/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+1960_Page_1.png" width="426" /></a></span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">2/2</span><br />
<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYmFU2ZpkwS5n9PMYYZ0OVScXb5wmQiKSQYd6HpceyuwRnoC-8owWLtI8WIU4BJz78GRF6t6x1wg2gAmJE6PyXPAcFz_O7ENdIq8auJysodt12jvs956MkQ4lJriJIaU2k8AZGK02-tfca/s1600/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+1960_Page_2.png" imageanchor="1"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgYmFU2ZpkwS5n9PMYYZ0OVScXb5wmQiKSQYd6HpceyuwRnoC-8owWLtI8WIU4BJz78GRF6t6x1wg2gAmJE6PyXPAcFz_O7ENdIq8auJysodt12jvs956MkQ4lJriJIaU2k8AZGK02-tfca/s640/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+1960_Page_2.png" width="424" /></a> </span></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-2590578030720929632017-07-19T18:04:00.004-07:002020-03-30T15:36:32.786-07:001959 IMO Problem 6 (CZS)<div dir="ltr" style="text-align: left;" trbidi="on">
<div dir="ltr" style="text-align: left;" trbidi="on">
proposed by Czechoslovakia<br />
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[solved , alternate solutions are welcome]</div>
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<span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Two planes, P and Q; intersect
along the line p: The point A is given in the plane P; and the point C in the
plane Q; neither of these points lies on the straight line p: Construct an
isosceles trapezoid ABCD (with AB parallel to CD) in which a circle can be
inscribed, and with vertices B and D lying in the planes P and Q respectively.</span></div>
</div>
</div>
<br />
<div style="text-align: right;">
<span style="font-size: small;">solved in aops</span><span style="font-size: large;"> </span><a href="https://artofproblemsolving.com/community/c6h54825p341533" style="font-size: x-large;" target="_blank">here</a><br />
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<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Δυο επίπεδα </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">P</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> και </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Q</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">τέμνονται κατά την ευθεία </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">p</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">. Το σημείο Α ανήκει στο επίπεδο </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">P</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">, ενώ το
σημειο </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">C</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> ανήκει στο
επίπεδο </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Q</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">, χωρίς κανένα
από τα </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">A</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">και </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">C</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">να ανήκει στην ευθεία </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">p</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">. Να κατασκευάσετε
ισοσκελές τραπέζιο </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">ABCD</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"><span style="mso-spacerun: yes;"> </span>με </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">AB</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">// </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">CD</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">, στο
οποίο να μπορεί να εγγραφεί ένας κύκλος και του οποίου οι κορυφές </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">B</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">και </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">D</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">να ανήκουν στα επίπεδα </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">P</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">και </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Q</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">, αντιστοίχως.</span></div>
</div>
</div>
</div>
<div dir="ltr" style="text-align: left;" trbidi="on">
<div style="text-align: right;">
<span style="color: red;">solved in pdf</span> <span style="font-size: large;"><a href="https://drive.google.com/file/d/1fIkgQ-yBHfW4o5gIebR2NeXS_V1QEsWc/view?usp=sharing" target="_blank">here</a></span><br />
<span style="font-size: medium;">(</span><span style="font-size: small;">with all 1959 geometry problems</span><span style="font-size: medium;"> <a href="https://drive.google.com/file/d/0B2qYu535vGeQWGprWGlmbW5GUkk/view?usp=sharing" style="font-size: x-large;" target="_blank">here</a>)</span><br />
<br /></div>
</div>
<div dir="ltr" style="text-align: left;" trbidi="on">
solved by Kostas Dortsios [translated in English inside the pdf]</div>
<div dir="ltr" style="text-align: left;" trbidi="on">
1/4</div>
<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTuhBE2_x754AP9KPoJZU7tkF04xq-N_EBpiKJENUTPajm4Lw7IlUAL_zIsI8pUVrdab2qaZWvod9TcyoZ5SFt2NbV7QK1UVkIWxTUAQRavYNbvzCUV8PfmFtOgn2bq3BngOB2zQhsOfEW/s1600/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+2%25281%2529_Page_1.png" imageanchor="1"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgTuhBE2_x754AP9KPoJZU7tkF04xq-N_EBpiKJENUTPajm4Lw7IlUAL_zIsI8pUVrdab2qaZWvod9TcyoZ5SFt2NbV7QK1UVkIWxTUAQRavYNbvzCUV8PfmFtOgn2bq3BngOB2zQhsOfEW/s640/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+2%25281%2529_Page_1.png" width="440" /> </a></div>
<div dir="ltr" style="text-align: left;" trbidi="on">
</div>
<div dir="ltr" style="text-align: left;" trbidi="on">
2/4</div>
<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXkxWBZDihR42yVx39kjHjvU9ojNPYZgw_1cIzlCvPgr0nRT9pPepbqeJeT7MkYPhk9hdadWiWCXw3bj0n8C7fV7IYFWibiVKE1DA30ZdnOhIKorpmxXZAKBTYW5tm00eS1KSJZQMqwp6I/s1600/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+2%25281%2529_Page_2.png" imageanchor="1"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgXkxWBZDihR42yVx39kjHjvU9ojNPYZgw_1cIzlCvPgr0nRT9pPepbqeJeT7MkYPhk9hdadWiWCXw3bj0n8C7fV7IYFWibiVKE1DA30ZdnOhIKorpmxXZAKBTYW5tm00eS1KSJZQMqwp6I/s640/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+2%25281%2529_Page_2.png" width="452" /></a> </div>
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3/4</div>
<div dir="ltr" style="text-align: left;" trbidi="on">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5J0PkSeDltcSx6tNNiCQb8xM5_HwM-e2GakxL7YFLNEGbNe3r2FUC__glTvJ-UtRaBn-r-dWTV6z1WiaxIA9IzheHYL9inW78ShCOWmTKjuvXOCM_H8XokNZW0E44vXona2V1k6E4FB4T/s1600/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+2%25281%2529_Page_3.png" imageanchor="1"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj5J0PkSeDltcSx6tNNiCQb8xM5_HwM-e2GakxL7YFLNEGbNe3r2FUC__glTvJ-UtRaBn-r-dWTV6z1WiaxIA9IzheHYL9inW78ShCOWmTKjuvXOCM_H8XokNZW0E44vXona2V1k6E4FB4T/s640/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+2%25281%2529_Page_3.png" width="452" /></a> </div>
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4/4</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgbohMQMDTZ38NAbvQm4S3NiFB2iRW-ShQZ-klo_ZyfCc9khnQ_qP8ld_aTLiw6vdOWjMcOUJh5HJuqSwVPLz_gryOr8_fqt4XZ734Z7YSbtf8VteCg2GFulxcQtlOWxAq-gE8OxPN_Qpf0/s1600/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+2%25281%2529_Page_4.png" imageanchor="1"><img border="0" height="640" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgbohMQMDTZ38NAbvQm4S3NiFB2iRW-ShQZ-klo_ZyfCc9khnQ_qP8ld_aTLiw6vdOWjMcOUJh5HJuqSwVPLz_gryOr8_fqt4XZ734Z7YSbtf8VteCg2GFulxcQtlOWxAq-gE8OxPN_Qpf0/s640/%25CE%25A3%25CF%2584%25CE%25B5%25CF%2581%25CE%25B5%25CE%25BF%25CE%25BC%25CE%25B5%25CF%2584%25CF%2581%25CE%25AF%25CE%25B1+2%25281%2529_Page_4.png" width="432" /></a> </div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-55481016725159020852017-07-19T18:03:00.004-07:002020-03-30T15:36:12.898-07:001959 IMO Problem 5 (ROM)<div dir="ltr" style="text-align: left;" trbidi="on">
proposed by Cezar Cosnita, Romania<br />
<div style="text-align: right;">
[solved, alternate solutions are welcome]</div>
<br />
<br />
<span lang="EN-US" style="font-family: "cmr12"; font-size: 12.0pt;">An arbitrary point M is selected in the interior of
the segment AB. The squares AMCD and MBEF are constructed on the same side of AB;
with the segments AM and MB as their respective bases. The circles
circumscribed about these squares, with centers P and Q; intersect at M and also
at another point N: Let N' denote the point of intersection of the straight lines
AF and BC:<br />
(a) Prove that the points N and N' coincide.<br />
(b) Prove that the straight lines MN pass through a fixed point S independent of the choice of M.<br />
(c) Find the locus of the midpoints of the segments PQ as M varies between A and B.</span>
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRLBmgg7T_6UoSRgvSvhBTCbLNFlf0mzIcBRaefKx6oDinaW9ILkaSx7oA8JiC6CNqobR5I3dmuuBC13-9TvyUVnC394gCuG-Mv4aktzy1g1IuSofBFfeBADIjeZJsL6TBR-BsQ7damwb-/s1600/1IMO5A.JPG" imageanchor="1"><img border="0" height="278" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjRLBmgg7T_6UoSRgvSvhBTCbLNFlf0mzIcBRaefKx6oDinaW9ILkaSx7oA8JiC6CNqobR5I3dmuuBC13-9TvyUVnC394gCuG-Mv4aktzy1g1IuSofBFfeBADIjeZJsL6TBR-BsQ7damwb-/s400/1IMO5A.JPG" width="400" /></a><br />
<div style="text-align: right;">
solved in aops <a href="https://artofproblemsolving.com/community/c6h54823p341530" target="_blank"><span style="font-size: large;">here</span></a><br />
<br />
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<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 0cm; margin-right: 0cm; margin-top: 6.0pt; text-align: justify;">
<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Θεωρούμε σημείο Μ εσωτερικό στο
ευθυγράμμο τμήμα ΑΒ . Τα τετράγωνα </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">AMCD</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> και </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">MBEF</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">κατασκευάζονται
προς το ίδιο μέρος του τμήματος </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">AB</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> με τα τμήματα ΑΜ
και ΜΒ ως βάσεις τους, αντιστοίχως</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">. Οι περιγεγραμμένοι κύκλοι των δυο τετραγώνων με
κέντρα </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">P</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">και </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Q</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">, αντίστοιχα, τέμνονται εκτός του
σημείου </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">M</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">και στο <span style="mso-spacerun: yes;"> </span>σημείο Ν.
Έστω Ν΄ το σημείο τομής των ευθειών </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">AF</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">και </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">BC</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">.</span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 0cm; margin-right: 0cm; margin-top: 6.0pt; text-align: justify;">
<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">(a) Να αποδείξετε οτι τα σημεία Ν
και Ν΄ συμπίπτουν. </span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 0cm; margin-right: 0cm; margin-top: 6.0pt; text-align: justify;">
<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">(b) Να αποδείξετε οτι οι ευθείες </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">MN</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> διέρχονται
από ένα σταθερό σημείο </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">S</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">ανεξάρτητο
από την επιλογή του </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">M</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">. </span></div>
<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 0cm; margin-right: 0cm; margin-top: 6.0pt; text-align: justify;">
<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">(c) Να βρείτε τον γεωμετρικό τόπο
των μέσων των τμημάτων </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">PQ</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">, καθώς το σημείο </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">M</span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;"> </span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">μεταβάλλεται<span style="mso-no-proof: yes;"> στο Α</span></span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">B</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">. </span></div>
<span style="color: red;">solved in pdf</span> <span style="font-size: large;"><a href="https://drive.google.com/file/d/0B2qYu535vGeQWGprWGlmbW5GUkk/view?usp=sharing" target="_blank">here</a></span><br />
<div style="text-align: right;">
<span style="font-size: x-small;">(<span style="background-color: whitesmoke; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;">with all 1959 geometry problems</span>)</span></div>
<br style="text-align: left;" /></div>
<span style="color: red;">my solution</span><br />
<br />
<span style="color: red;"><span style="color: black;">[english / greek] </span></span><br />
(a) <br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzu5FrTyuWREjYHWQbzjwX1wegfDMBNh6ONbv3UxuZJiIqD5El37rVSjJhbs4GVTejsHPxWfFKYnFWg0Y4LfU0LrTUw3isnT5pjHpVdr4v0HWz18IPm1GmVIOHTF679Wh1waFL7GyJeBhf/s1600/imo+1959+part+a.png" imageanchor="1"><img border="0" height="282" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzu5FrTyuWREjYHWQbzjwX1wegfDMBNh6ONbv3UxuZJiIqD5El37rVSjJhbs4GVTejsHPxWfFKYnFWg0Y4LfU0LrTUw3isnT5pjHpVdr4v0HWz18IPm1GmVIOHTF679Wh1waFL7GyJeBhf/s640/imo+1959+part+a.png" width="640" /></a><br />
<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHBA8VnXcQBZ2_pZL1a0fGsZZEv4EFIQciR55REFaoDRRMjQMpOqDetnawQZ1x0adeFsvyTi5bv3pJWX8v9bvaU5TvVKyDcRL5SvhzKt7POmV5Zv7j4ri7wawxaviNTUzXqfhCSq-FRK7i/s1600/imo+1959+part+a+greek.png" imageanchor="1"><img border="0" height="306" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjHBA8VnXcQBZ2_pZL1a0fGsZZEv4EFIQciR55REFaoDRRMjQMpOqDetnawQZ1x0adeFsvyTi5bv3pJWX8v9bvaU5TvVKyDcRL5SvhzKt7POmV5Zv7j4ri7wawxaviNTUzXqfhCSq-FRK7i/s640/imo+1959+part+a+greek.png" width="640" /></a><br />
(b)<br />
<br />
<br />
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<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxM30JGTbMOqviUD4u7kmbjwMhObgGLjbdxbDxtQ_X70oO6EVU2RMaTpqfDTev6v_J-zJ0pPx7qR5yESstHh6jMwQbeHxd6qH-nA9sYO-mob_xi8ouGysMvNtxjtxhbhzx9zdMw-9ucHSs/s1600/imo+1959+part+b.png" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" height="308" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjxM30JGTbMOqviUD4u7kmbjwMhObgGLjbdxbDxtQ_X70oO6EVU2RMaTpqfDTev6v_J-zJ0pPx7qR5yESstHh6jMwQbeHxd6qH-nA9sYO-mob_xi8ouGysMvNtxjtxhbhzx9zdMw-9ucHSs/s640/imo+1959+part+b.png" width="640" /></a></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3RRvUmyBq6AvkzJw4olxquBkyGNcM8Lp8Nkv5O0ZonnUa4TUym6lUk1FHP_mfQe4p7KF8_Cr-DepANS7z4k9TCB6xMPryJSyO5ZPldMDTg5_xHEtY7At_Hd8653nJ5aiqpZcsKvwwTIK4/s1600/imo+1959+part+b+greek.png" imageanchor="1"><img border="0" height="306" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj3RRvUmyBq6AvkzJw4olxquBkyGNcM8Lp8Nkv5O0ZonnUa4TUym6lUk1FHP_mfQe4p7KF8_Cr-DepANS7z4k9TCB6xMPryJSyO5ZPldMDTg5_xHEtY7At_Hd8653nJ5aiqpZcsKvwwTIK4/s640/imo+1959+part+b+greek.png" width="640" /></a><br />
(c)<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEimck9ArsEmEPQqVRI2J_sBXURr8qoC9lnzPaIRQIodI2KsxE6FaWtyKhIcd_J1Nhx527av1KjJo-_PsDCdnIQtW1qn3X6vtK4yLfRMfL50YYTjoc4-Cr_4_jqi_2SK5jV6PlDqMA0S9kWq/s1600/imo+1959+part+c.png" imageanchor="1"><img border="0" height="306" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEimck9ArsEmEPQqVRI2J_sBXURr8qoC9lnzPaIRQIodI2KsxE6FaWtyKhIcd_J1Nhx527av1KjJo-_PsDCdnIQtW1qn3X6vtK4yLfRMfL50YYTjoc4-Cr_4_jqi_2SK5jV6PlDqMA0S9kWq/s640/imo+1959+part+c.png" width="640" /> </a><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjy5qFUCCYrAVxgjgsJ-EtjfrIQd9k1nv2g1LdZ97pw8jqkGeO2nm-v8sn1yS8r_RbE3eMLJiVYJ3sgxoyQqPfznmUnhU1D8UcZ_kq5WsL43hyS-VyxrSQMb3UdhroayDDUGx1P8JnuKsTt/s1600/imo+1959+part+c+greek.png" imageanchor="1"><img border="0" height="306" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjy5qFUCCYrAVxgjgsJ-EtjfrIQd9k1nv2g1LdZ97pw8jqkGeO2nm-v8sn1yS8r_RbE3eMLJiVYJ3sgxoyQqPfznmUnhU1D8UcZ_kq5WsL43hyS-VyxrSQMb3UdhroayDDUGx1P8JnuKsTt/s640/imo+1959+part+c+greek.png" width="640" /> </a><br />
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(a) solution by Giannopoulos Theodosis<br />
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_IF9hBKbfxTy3hoamLMSS4Dc5soW2n-XNcC0AzAupNkmVa0MfpC8r5Xk99Yf18-gNMgECNOonqtcNAHy8mTrinkfPbwRXofT2koqLU76NGAL4ItL_JN-nU-OGkZ3WWZendbfxvZDBht10/s1600/imo+1959+part+a2.png" imageanchor="1"><img border="0" height="281" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_IF9hBKbfxTy3hoamLMSS4Dc5soW2n-XNcC0AzAupNkmVa0MfpC8r5Xk99Yf18-gNMgECNOonqtcNAHy8mTrinkfPbwRXofT2koqLU76NGAL4ItL_JN-nU-OGkZ3WWZendbfxvZDBht10/s640/imo+1959+part+a2.png" width="640" /></a><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipY2AlzNuQB92wFK66p_YL2dTw1Z9Sup5AIKoa345D39bkKzVgTQ5Wq_Ahv_QIIbRFl7moWuBj94Bz8vDFPHgGXbMM6EujL_QiZgqwVyHTmHPnXVDnQR6mVs6u_eughRfSboCMl4TZAQBm/s1600/imo+1959+part+a2+greek.png" imageanchor="1"><img border="0" height="306" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEipY2AlzNuQB92wFK66p_YL2dTw1Z9Sup5AIKoa345D39bkKzVgTQ5Wq_Ahv_QIIbRFl7moWuBj94Bz8vDFPHgGXbMM6EujL_QiZgqwVyHTmHPnXVDnQR6mVs6u_eughRfSboCMl4TZAQBm/s640/imo+1959+part+a2+greek.png" width="640" /></a></div>
parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0tag:blogger.com,1999:blog-7236457855785919045.post-62555089351114396312017-07-19T18:00:00.003-07:002020-03-30T15:35:57.050-07:001959 IMO Problem 4 (HUN)<div dir="ltr" style="text-align: left;" trbidi="on">
proposed by Hungary<br />
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[solved, alternate solutions are welcome]</div>
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Construct a right triangle with given hypotenuse c such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.<br />
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<span style="font-size: small;">solved in aops</span><span style="font-size: large;"> </span><a href="https://artofproblemsolving.com/community/c6h54821p341522" style="font-size: x-large;" target="_blank">here</a></div>
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<div class="MsoNormal" style="line-height: normal; mso-margin-bottom-alt: auto; mso-margin-top-alt: auto; text-align: justify;">
<span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">Να κατασκευάσετε ορθογώνιο τρίγωνο με δεδομένη
υποτείνουσα </span><span lang="EN-US" style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">c</span><span style="font-family: "times new roman" , "serif"; font-size: 12.0pt;">, έτσι ώστε
η διάμεσος που αντιστοιχεί στην υποτείνουσα να ισούται με τον γεωμετρικό μέσο
των δυο κάθετων πλευρών του τριγώνου.</span></div>
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<span style="color: red;">solved in pdf</span> <span style="font-size: large;"><a href="https://drive.google.com/file/d/0B2qYu535vGeQWGprWGlmbW5GUkk/view?usp=sharing" target="_blank">here</a></span><br />
<span style="font-size: x-small;">(<span style="background-color: whitesmoke; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;">with all 1959 geometry problems</span>)</span></div>
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<b><span style="color: red;">my solution</span></b><br />
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[english] <br />
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvB5eTpHQo9LeBRyTryvZLpiGzhYr0UsEWiiglv0SQpor6iV5efJBC5k4CcBZYCCfkpsAECndXZp_yX7l77-p9DGtX-GbgmPp516Xsbi2DRDqVPsBiNrBDymxxWjWBc4a1FqEoB_TF8hmn/s1600/imo1959+no4.jpg" imageanchor="1"><img border="0" height="316" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhvB5eTpHQo9LeBRyTryvZLpiGzhYr0UsEWiiglv0SQpor6iV5efJBC5k4CcBZYCCfkpsAECndXZp_yX7l77-p9DGtX-GbgmPp516Xsbi2DRDqVPsBiNrBDymxxWjWBc4a1FqEoB_TF8hmn/s640/imo1959+no4.jpg" width="640" /></a><br />
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[greek]</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6YgH-g_wuc1fiZpUS9mEH8_CEtW2SgfH32-67TUFtdkvr2Ur5cer_PmJr1iJm_DaJU8_3H59i3aMe0WIGzU9hC3dJRL8PaFNesl_MD0xZcdce86J5d2cGWPe66-fAd-QYhSICfNIdtnt0/s1600/imo1959+no4+greek.png" imageanchor="1"><img border="0" height="316" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh6YgH-g_wuc1fiZpUS9mEH8_CEtW2SgfH32-67TUFtdkvr2Ur5cer_PmJr1iJm_DaJU8_3H59i3aMe0WIGzU9hC3dJRL8PaFNesl_MD0xZcdce86J5d2cGWPe66-fAd-QYhSICfNIdtnt0/s640/imo1959+no4+greek.png" width="640" /></a></div>
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parmenides51http://www.blogger.com/profile/16912957097789786696noreply@blogger.com0